1
A solution is a triple that satisfies all three equations
(1, 2, 3) means x = 1, y = 2, z = 3, together. A triple that satisfies two of the three equations is not a partial solution; it is not a solution.
Lesson · Systems and Matrices · Systems
Systems in Three Variables
A solution is now an ordered triple, and each equation describes a plane. The method is elimination used twice: remove one variable from two different pairs of equations, solve the two-variable system that remains, and back-substitute.
x + y + z = 6 E1
2x − y + 3z = 9 E2 E2 − 2E1 : −3y + z = −3
−x + 2y + 4z = 15 E3 E3 + E1 : 3y + 5z = 21
add : 6z = 18 → (1, 2, 3)
Eliminate x twice, then solve what is left
Core ideas
What this lesson establishes
Nothing new is needed. What is new is the bookkeeping, and one decision that is easy to get wrong.
2
Eliminate the same variable from two different pairs
This is the step that goes wrong. Removing x from one pair and y from another leaves two equations containing three unknowns between them, and nothing has been reduced.
3
The three cases return, with more geometry behind them
One solution, none, or infinitely many. In three variables the infinite case can be a line or a plane, and the algebra reports which by how many variables are left free.
The material
Three equations, three unknowns
The reduction to a two-variable system, the geometry of each outcome, and the bookkeeping that keeps it honest.
Definition · Linear equation in three variables
An equation that can be written ax + by + cz = d with a, b, c not all zero. Its graph is a plane in space. A solution of a system of such equations is an ordered triple (x, y, z) satisfying every equation in the system.
The method: choose a variable, eliminate it from two different pairs of equations, and you are left with two equations in the remaining two unknowns — the situation of “Solving by Elimination”. Solve that, then back-substitute into any original equation to recover the variable that was eliminated.
Theorem · The reduction finds every solution and no others
Suppose x is eliminated from E₁ and E₂ to give A, and from E₁ and E₃ to give B, where the coefficient of x in E₁ is not zero. Then every solution of the original system satisfies A and B; and for each pair (y, z) satisfying A and B, equation E₁ determines exactly one x, and the resulting triple solves the original system.
The theorem is “Solving by Elimination”'s theorem used twice, which is why every elimination must keep E₁ intact and combine it with something else. It also explains the requirement that the two new equations come from different pairs: two combinations of E₁ and E₂ alone tell you only what E₁ and E₂ already said, and E₃ never enters the calculation.
What the three planes are doing
- Exactly one solution. The three planes have a single common point, the way two walls and a floor meet in a corner.
- No solution. Two of the planes may be parallel and distinct; or the three may meet in pairs along three parallel lines, like the sides of a triangular prism, with no point on all three.
- Infinitely many. The planes may share a whole line — one free variable — or all three equations may describe the same plane, in which case two variables are free.
That picture is worth having and is not worth relying on. Nobody reads a triple off a drawing of three planes, and the prism case in particular is invisible until the algebra produces a false statement. The algebra is the arbiter here even more completely than in “Systems of Two Linear Equations”.
Example · A third equation that adds nothing
Replace E₃ above by 3x + 4z = 15, which is E₁ + E₂. Eliminating x twice now ends in 0 = 0, and the solution set is a line: (9 − 4t, t, 3t − 3).
At t = 1 the triple is (5, 1, 0), and 5 + 1 + 0 = 6, 10 − 1 + 0 = 9, 15 + 0 = 15. At t = 2 it is (1, 2, 3), the solution the original system had on its own. One equation of the three was never carrying any information.
A false statement means the opposite. x + y + z = 3 together with 2x + 2y + 2z = 7 gives 0 = 1 on the first elimination, and the system is inconsistent no matter what the third equation says.
Bookkeeping
- Number the equations E1, E2, E3 and write beside every new equation the combination that produced it — E2 − 2E1. Without that record an error cannot be located, only redone.
- Keep the columns aligned and write a 0 for any variable missing from an equation. A missing term is information, not an absence.
- Do the back-substitution in the reverse order of the elimination: the last variable found first, then the second, then the one that left first.
This bookkeeping is the entire content of “Matrices and Row Operations”. Once the columns are aligned and every coefficient has a fixed position, the letters x, y, z are doing no work, and deleting them costs nothing.
Interactive
Three equations, three unknowns
Gaussian elimination on the augmented matrix, then back substitution
Method
Solving a three-variable system by hand
Four moves. Write the plan down before doing any arithmetic.
1
Number the equations and put them in standard form
x, y, z in that order in every equation, constants on the right, and a 0 written in for any variable that does not appear. Alignment errors account for most of the wrong answers in this lesson.
2
Choose the variable that is cheapest to eliminate
Look for a coefficient of 1 or −1, or two coefficients that are already opposites. Ten seconds spent here saves a page.
3
Eliminate it from two different pairs
E1 with E2, and E1 with E3, is the standard choice and uses all three equations. Label each result with the combination that made it.
4
Solve the two-variable system, back-substitute, check in all three
The third check is not ceremonial: it is the only test of the arithmetic you did during the elimination itself.
Three variables
Score 0 / 0Choose the answer you can justify.
A three-variable system, start to finish
x has coefficients 1, 2, −1, so it is the cheap variable to remove: E2 − 2E1 and E3 + E1, both against E1.
E1 x + y + z = 6
E2 2x − y + 3z = 9
E3 −x + 2y + 4z = 15
E2 − 2E1 : (2 − 2)x + (−1 − 2)y + (3 − 2)z = 9 − 12
A : −3y + z = −3
E3 + E1 : (−1 + 1)x + (2 + 1)y + (4 + 1)z = 15 + 6
B : 3y + 5z = 21
A + B : 6z = 18 → z = 3
A : −3y + 3 = −3 → −3y = −6 → y = 2
E1 : x + 2 + 3 = 6 → x = 1
check E1 : 1 + 2 + 3 = 6 ✓
E2 : 2 − 2 + 9 = 9 ✓
E3 : −1 + 4 + 12 = 15 ✓
solution (1, 2, 3)
Both eliminations removed x, which is the point of the whole exercise. E2 + 2E3 would also have been acceptable — a different pair, the same variable, giving 3y + 11z = 39. What is not acceptable is removing a different variable the second time: 2E2 + E3 gives 3x + 10z = 33, a perfectly true equation in x and z, which together with A in y and z involves three unknowns and leaves the system exactly as large as it was.
The back-substitution runs backwards through the work: z from the sum, y from A, x from E1. Substituting into E2 or E3 instead is equally valid and slightly more arithmetic; what matters is that all three are used for the final check.
Faded rehearsal: for x + y + z = 4, 2x − y + z = 1, −x + 3y + 2z = 9, the plan is supplied: form E2 − 2E1 and E3 + E1, so x leaves both new equations. Carry out those two combinations, solve the resulting system, and test the triple in all three originals.
Common mistakes
1
Eliminating a different variable the second timeThe two new equations must be in the same two unknowns. This single error turns a five-minute exercise into an afternoon, and it is invisible until the two-variable system refuses to solve.
2
Combining the same pair twiceA second combination of E1 and E2 alone follows from the first and says nothing new. All three equations must take part, or the information in the one left out never reaches the answer.
3
Checking only in the equations you finished withThe triple satisfies A and B by construction, so testing it there proves nothing about the elimination that produced them. Substitute into E1, E2 and E3 as they were given.
Check your understanding
✓
Finish when the plan is written before the arithmetic starts.
State which variable is leaving and which two pairs will remove it, then execute in exact fractions and check in all three original equations. A three-variable system solved by improvising comes out right about half the time; solved to a plan it is routine, and the plan is precisely what “Gaussian Elimination” automates.
Note: From here to “Inverse Matrices” this is the same arithmetic with the letters deleted. “Matrices and Row Operations” writes x + y + z = 6 as the row 1 1 1 | 6 and E2 − 2E1 as R2 − 2R1; nothing about the numbers changes, which is exactly why it is worth having done one of these by hand first.